\section{1.1. Differential operators on an affine variety} 
\begin{frame}[allowframebreaks]{1.1. }

\vspace{-0.4cm}

1.1. Let first $X$ be affine. 

The ring $\mathcal{D}(X)$ of (algebraic) differential operators on $X$ is the ring of operators on $\mathcal{O}(X)$, acting by multiplication, and by $\Theta_X(X) = \mathrm{Der}_\mathbb{C}(\mathcal{O}(X), \mathcal{O}(X))$. 

We have therefore in $\mathcal{D}(X)$ the rule
\begin{equation}
[\partial, f] = \partial f \quad (f \in \mathcal{O}(X), \partial \in \mathcal{D}(X))
\end{equation}

We shall soon deduce from V (F. Ehlers. The Weyl Algebra) that $\mathcal{D}(X)$ is noetherian, coherent, of global homological dimension equal to $\dim X$ (see 1.9).

If $Y$ is open affine, then $\Theta(Y)$ is generated over $\mathcal{O}(Y)$ be the restrictions of the elements in $\Theta(X)$, i.e. $\Theta(Y) = \mathcal{O}(Y) \otimes_{\mathcal{O}(X)} \Theta(X)$, whence also
\begin{equation}
\mathcal{D}(Y) = \mathcal{O}(Y) \otimes_{\mathcal{O}(X)} \mathcal{D}(X)
\end{equation}

There exists then {\color{red}a unique quasi-coherent $\mathcal{O}_X$-module on $X$}, denoted $\mathcal{D}_X$, whose sections over $Y$ open affine in $X$ is $\mathcal{D}(Y)$.


\newpage

{\color{red}Notebook.} 

1. Please give an example to illustrate this procedure, first we have $\mathcal{D}(X)$, then for each affine open $Y$ we have $\mathcal{D}(Y)$, and finally we glue them together to get the sheaf $\mathcal{D}_X$. (Consider $X=\mathbb{A}^1_\mathbb{C}$ and $Y=X-\{0\}$.)

2. Here $X$ is an affine algebraic variety. Is it possible that an open subset $Y$ of $X$ not affine? How do we define $\mathcal{D}_X(Y)$ in this case? (Consider $X=\mathbb{A}^2_\mathbb{C}$ and $Y=X-\{(0,0)\}$.)



\newpage

1K.  

To illustrate the procedure of constructing the sheaf $\mathcal{D}_X$ from the rings $\mathcal{D}(Y)$ for affine open subsets $Y$ of $X$, let's consider a simple example. 

Suppose $X = \mathbb{A}^1_\mathbb{C}$, the affine line over the complex numbers. 

The ring of regular functions on $X$ is simply the polynomial ring in one variable, $\mathcal{O}(X) = \mathbb{C}[x]$. 

The derivations $\Theta(X)$ are generated by the single derivation $\partial_x = \frac{d}{dx}$, which acts on $\mathcal{O}(X)$ by differentiation with respect to $x$.

Step 1: Construct $\mathcal{D}(X)$

The ring of differential operators $\mathcal{D}(X)$ on $X$ consists of all finite sums of compositions of multiplication by elements of $\mathcal{O}(X)$ and the derivation $\partial_x$. 

Thus, $\mathcal{D}(X)$ is the Weyl algebra in one variable, denoted as $A_1(\mathbb{C})$ or $\mathbb{C}\langle x, \partial_x \rangle$, subject to the relation $[\partial_x, x] = 1$. 

This means every element of $\mathcal{D}(X)$ can be written uniquely as a finite sum
\[
P = \sum_{i,j} c_{ij} x^i \partial_x^j,
\]
where $c_{ij} \in \mathbb{C}$.

Step 2: Consider an Open Affine Subset $Y$

Now, take any non-empty open affine subset $Y$ of $X$. 

Since $X = \mathbb{A}^1_\mathbb{C}$, any such $Y$ is itself isomorphic to $\mathbb{A}^1_\mathbb{C}$ minus a finite number of points. 

For simplicity, let's choose $Y = X - \{0\}$, i.e., the affine line punctured at the origin. 

The ring of regular functions on $Y$ is then $\mathcal{O}(Y) = \mathbb{C}[x, x^{-1}]$, the localization of $\mathbb{C}[x]$ at $x$.

The ring of differential operators $\mathcal{D}(Y)$ on $Y$ is similarly constructed but now includes terms involving negative powers of $x$ due to the presence of $x^{-1}$ in $\mathcal{O}(Y)$. 

However, since $\mathcal{D}(Y)$ must be compatible with $\mathcal{D}(X)$, it follows that $\mathcal{D}(Y) = \mathcal{O}(Y) \otimes_{\mathcal{O}(X)} \mathcal{D}(X)$, meaning we can still use the same generators and relations as in $\mathcal{D}(X)$ but now allow coefficients from $\mathcal{O}(Y)$.

Step 3: Glue Together to Form $\mathcal{D}_X$

Given the compatibility between $\mathcal{D}(X)$ and $\mathcal{D}(Y)$ for any affine open subset $Y$ of $X$, we can construct the sheaf $\mathcal{D}_X$ on $X$ by specifying its sections on each affine open subset. 

Specifically, if $U$ is any open subset of $X$, then $\mathcal{D}_X(U)$ consists of those collections $(s_Y)_{Y \subseteq U}$ where $s_Y \in \mathcal{D}(Y)$ for each affine open $Y \subseteq U$, and these sections agree on overlaps, i.e., $s_Y|_{Y \cap Y'} = s_{Y'}|_{Y \cap Y'}$ for any two affine opens $Y, Y' \subseteq U$.

In our example, since $\mathcal{D}(Y)$ for any open affine $Y \subseteq X$ is determined by the action of $\mathcal{D}(X)$, gluing them together to form $\mathcal{D}_X$ is straightforward. 

The resulting sheaf $\mathcal{D}_X$ assigns to each open set $U$ the ring of differential operators that act on the ring of regular functions $\mathcal{O}_X(U)$.

This construction illustrates how the global object $\mathcal{D}_X$ is assembled from local pieces $\mathcal{D}(Y)$, ensuring that the resulting sheaf of rings satisfies the necessary compatibility conditions across different affine open subsets of $X$.


\newpage 

2K. 

% (1) $\mathcal{D}_X(Y)$ is generated by $\mathcal{O}_X(Y)$ and $\mathrm{Der}_\mathbb{C}(\mathcal{O}_X(Y))$.

% (2) $\mathcal{D}_X(Y)$ is defined theaf-theoretically by $\mathcal{D}_X$. 

Yes, it is possible for an open subset $Y$ of an affine algebraic variety $X$ to not be affine. 

A classic example is the punctured affine plane: let $X = \mathbb{A}^2_\mathbb{C} = \mathrm{Spec}\,\mathbb{C}[x, y]$, which is affine. 

Let $Y = X - \{(0,0)\}$, the complement of the origin. 

Then $Y$ is an open subset of $X$, but it is not affine.


To see this, note that the ring of regular functions on $Y$ is still $\mathbb{C}[x, y]$, because any regular function on $Y$ extends uniquely to a regular function on $X$ (by Hartogs' theorem in algebraic geometry, valid in dimension $\ge 2$). 

So $\mathcal{O}(Y) = \mathbb{C}[x, y] = \mathcal{O}(X)$. 

If $Y$ were affine, then $Y$ would be isomorphic to $\mathrm{Spec}\,\mathcal{O}(Y) = \mathrm{Spec}\,\mathbb{C}[x, y] = X$, which is not true since $Y \subsetneq X$. 

Hence, $Y$ is not affine.


How do we define $\mathcal{D}_X(Y)$ when $Y$ is not affine?

Even when $Y$ is not affine, we define the sections $\mathcal{D}_X(Y)$ of the sheaf $\mathcal{D}_X$ on $Y$ as the ring of differential operators on the sheaf $\mathcal{O}_X|_Y$.

More precisely:

The sheaf $\mathcal{D}_X$ is a sheaf of rings on $X$, constructed so that on every affine open subset $U \subseteq X$, we have {\color{red}$\mathcal{D}_X(U) = \mathcal{D}(U)$}, the ring of algebraic differential operators on $U$.

For an arbitrary open subset $Y \subseteq X$ (affine or not), $\mathcal{D}_X(Y)$ is defined as the ring of global sections of the sheaf $\mathcal{D}_X$ over $Y$, i.e., {\color{red}$\mathcal{D}_X(Y) = \Gamma(Y, \mathcal{D}_X)$}.

Concretely, an element of $\mathcal{D}_X(Y)$ is a collection of differential operators defined on affine open subsets of $Y$, compatible under restriction.

Alternatively, one can define $\mathcal{D}_X(Y)$ as the subring of $\mathrm{End}_\mathbb{C}(\mathcal{O}_X(Y))$ consisting of operators that are {\color{blue}differential operators of some finite order} on $\mathcal{O}_X(Y)$, meaning they satisfy the usual inductive definition: an operator $P$ is a differential operator of order $\le m$ if $[P, f]$ is a differential operator of order $\le m-1$ for all $f \in \mathcal{O}_X(Y)$, with order 0 operators being multiplication operators.

In the case where $Y$ is not affine, $\mathcal{D}_X(Y)$ is not simply the ring of differential operators on the ring $\mathcal{O}_X(Y)$ in the abstract algebraic sense (i.e., generated by $\mathcal{O}_X(Y)$ and $\mathrm{Der}(\mathcal{O}_X(Y))$), because that ring might not capture the correct sheaf-theoretic behavior. 

Instead, $\mathcal{D}_X(Y)$ is defined sheaf-theoretically as the sections of the sheaf $\mathcal{D}_X$ over $Y$.

So, in summary:

For any open subset $Y \subseteq X$, whether affine or not, $\mathcal{D}_X(Y) := \Gamma(Y, \mathcal{D}_X)$, where $\mathcal{D}_X$ is the sheaf of differential operators on $X$, uniquely determined by its values on affine opens via $\mathcal{D}_X(U) = \mathcal{D}(U)$ for affine $U$.



\end{frame}

